husoski
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Assuming they don't intersect, the difference will be an extreme point of:
f(x) = 101x^6 - 404x^2 + 311
f'(x) = 606x^5 - 808x = 202 x(3x^4 - 4)
The real roots of f' are x = 0 and x = ± ⁴√(4/3).
The second derivative is:
f"(x) = 3030x^4 - 808
That's clearly negative at x=0, meaning f(0) is a local maximum, and just as clearly positive when |x| > 1, so x = ± ⁴√(4/3) are local minima.
Since f(x) is symmetrical (only even powers) you only need to look at the positive value x=⁴√(4/3). If f(x) > 0 there, then there can't be any root of f between 0 and that x value (else Rolle's theorem says that f' would be zero somewhere in-between, and we've just shown that x and 0 are the only non-negative roots of f'.)
f(⁴√(4/3)) = 101(4/3)^(3/2) - 404(4/3)^(1/2) + 311
= 101*√(4/3)*[(4/3) - 4] + 311
= 311 - (808/3)√(4/3)
That second term is computationally about 310.999345, so the curves don't touch and that value is positive and equal to the exact closest *vertical* distance.
Finding the closest *geometric* distance between the graphs is a much harder problem.
J
I decided to work it out. I contrived the problem, but had not worked out the answer when I posted.
May as well minimize the squared distance between (X, 101X^6) and (x, 404x^2-311) where X and x are independent. Writing down that squared distance and taking partial derivatives, gives the system of two polynomial equations in the two unknowns x, X:
61206X^11 - 244824X^5*x^2 + 188466X^5 + X - x = 0
-81608X^6*x + 326432x^3 - X - 251287x = 0
Note that the first equation is quadratic in x.
One could use a multivariate Newton, say, once you have an approximate solution.
Or throw it to wolframalpha, see
http://www.wolframalpha.com/input/?i=solve+%7B61206X%5E11+-+244824X%5E5*x%5E2+%2B+188466X%5E5+%2B+X+-+x+%3D+0,++-81608X%5E6*x+%2B+326432x%5E3+-+X+-251287x+%3D+0%7D,+%7BX,x%7D
You will need to click both "more roots" and "more digits". For some reason wolfram only gave me X.
Ultimately using Maple I get the points
(-1.074570120419334688501768725, 155.4998362505805418102621281), (-1.074570874802836765361492622, 155.4998362497116899085332346)
one on each of the two graphs, and the distance between them as
0.00000075438400242
smaller than husoski's vertical distance which was
0.006549965215881844